How do I find the average of a section of a curve? I've done some research already and discovered that the formula I need to do this is
$$\frac{1}{b-a}\int_a^b f(x)\;dx$$
With $a$ and $b$ being the start and end points of the section of curve I want to get the average of.
What I need to know is how I translate a maths $y = ???$ (featuring $x$) function into $f(x)\;dx$, so I can actually work out what the formula I need is.
I gather that $f(x)$ means "a formula of a curve" and that $f(x)\;dx$ means a formula for figuring out the area beneath the curve of the given $f(x)$, but I haven't been able to find a manual for how to translate a given $f(x)$ into the $f(x)\;dx$ that gives it's area.
Can somebody help me with this?
 A: Well, if you have an equation $y=x^2$ then take $f(x)=x^2$, meaning that just take $y=f(x)$. If you are not familiar with calculus and insist on finding the average this way, then there isn't a good way for me to explain the dx part. Just write the dx in the integral. Also do you know how to compute the integral $\int_a^b f(x)dx$?
So if you want to find the average value of $y=x^2$ between $x=7$ and $x=9$, then it is
Average = $\frac{1}{9-7}\int_7^9 x^2dx = \frac{1}{2}\cdot\frac{x^3}{3}|_7^9 = \frac{729}{6}-\frac{343}{6}=\frac{386}{6}=64 \frac{1}{3}.$
Also taking y=f(x) is not always possible. The equation must be y = a function of x and x alone. There shouldn't be any y's on the right hand side and the right hand side must pass the vertical line test.
A: No offense, but your math seems to be far behind of what you wish to achieve with it. If you can live with a value near the average I suggest you take some values between $a$ and $b$, like say
$x_i = a + i\frac{b-a}{n}$
for some fixed numer $n$, calculate the function at these values ($f(x_i)$), sum them all up and divide by $n$. The higher your $n$ the more accurate your average will be.
The big-stretched-out-"S" as you call it (integral) does exactly that, but with $n$ going to infinity. If you want to really understand this - and it is a really beautiful theory with some unimaginable results, everybody should see it - you should read some of the standard literature on calculus. This might also be a promising start:
http://en.wikipedia.org/wiki/Integral_calculus
To interpret the integral as the area between the curve and the x axis is only one way to see it. In my view this way does not help a lot in this case.
A: we usually don't sum all the approximating rectangles, but perform that task indirectly by invoking the fundamental theorem of calculus.
we suppose that
$$F(t)=\int_a^t f(x)dx$$
represents the area under the curve (or more correctly, the net area between the curve and the x-axis) of $f$ where $a\le x\le t$.
Now consider the derivative of A (the function whose values represent the slope of $A$)
$$F'(t)=\lim_{h\to 0} \frac{F(t+h)-F(t)}{h}$$
$$F'(t)=\lim_{h\to 0} \frac{\int_a^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$$
$$F'(t)=\lim_{h\to 0} \frac{\int_a^{t} f(t)dt+\int_t^{t+h} f(t)dt-\int_a^{t} f(t)dt}{h}$$
$$F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$$

Now think about what would happen to the shaded region to the right (between $t$ and $t+h$) were $h$ to become very small. The left and right sides of the region would become closer to each other in length and the region would resemble a rectangle with a width of $h$ and a height of $f(t)$.
thus
$$F'(t)=\lim_{h\to 0} \frac{\int_t^{t+h} f(t)dt}{h}$$
$$F'(t)=\lim_{h\to 0} \frac{hf(t)}{h}$$
$$F'(t)=f(t)$$
Put another way, we say that the area function ($F$) is some anti-derivative of $f$.
For example, if we know that the antidervivative of $x^2$ is $x^3/3+C$ (we have an unknown constant because parallel curve have the same slope), we know that the area between where x ranges between $1$ and $2$ would be $2^3/3+C$, which is a pretty useless result. To fix the problem we note that area of a region is just the sum of the areas of its components.

In the second graphic, the total area under $f$ between $c$ and $b$ is just the sum of the blue and green areas. In the language of integrals,
$$\int_c^bf(x)dx = \int_c^af(x)dx + \int_a^bf(x)dx$$
thus
$$\int_a^bf(x)dx = \int_c^bf(x)dx - \int_c^af(x)dx$$
So to solve the area problem, we need to evaluate 2 anti-derivatives, not one. In our example
$$\int_1^2 x^2dx = (2^3/3+C)-(1^3/3+C)=(8-1)/3=7/3$$
Notice how nicely the $C$s cancel out.
By the way, in single variable calculus, you can think of $dx$ as an infinitely small non-zero positive $\Delta x$. Note that $d$ and $\Delta$ are the Latin and Greek representations for the first sound in the word difference.
